Question

find the work done if f=3i +4j-5k and displacement s=5i+4j+5k.Also find the angle between force and displacement​

Answer 1

Answer:

$\text{work done is 6}$

$\text{Angle between force and displacement is}\:cos^{-1}(\frac{\sqrt3}{5\sqrt{11}})$

Explanation:

$\text{Given:}$

$\text{Force:}\:\vec{f}=3\vec{i} +4\vec{j}-5\vec{k}$

$\text{displacement:}\:\vec{s}=5\vec{i} +4\vec{j}+5\vec{k}$

$\text{Work done}=\vec{f}.\vec{s}$

$\text{Work done}=(3\vec{i} +4\vec{j}-5\vec{k}).(5\vec{i} +4\vec{j}+5\vec{k})$

$\text{Work done}=(3)(5)+(4)(4)+(-5)(5)$

$\text{Work done}=15+16-25$

$\text{Work done}=6$

$\text{Angle between force and displacement}$

$cos\theta=\frac{\vec{f}.\vec{s}}{|\vec{f}|\:|\vec{s}|}$

$=\frac{6}{\sqrt{9+16+25}\:\sqrt{25+16+25}}$

$=\frac{6}{\sqrt{50}\:\sqrt{66}}$

$=\frac{\sqrt6}{\sqrt{50}\:\sqrt{11}}$

$=\frac{\sqrt3}{\sqrt{25}\:\sqrt{11}}$

$cos\theta=\frac{\sqrt3}{5\sqrt{11}}$

$\theta=cos^{-1}(\frac{\sqrt3}{5\sqrt{11}})$

Answer 2

Cos.teta=f.d/fd

F.d=(3i+4j-5k)(5i+4j+5k)=15+16-15=16

Fd=(root9+16+25)(root25+16+9)

=root50.root50

=50

Cos.teta=16/32

teta=cos^-1.0.32